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  • In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of A meets DC in E. AE and BC produced meet at F. Find the length of CF.

In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of \angle A meets DC in E. AE and BC produced meet at F. Find the length of CF.

Answers (1)

Answer: 4cm
 

Solution.
 

Given: ABCD is a parallelogram in which AB = 10 cm  and AD = 6 cm.
Construction: In parallelogram ABCD, draw the bisector of \angle A  which meets DC in point E. 
Produce AE and BC so that they meet at point F.
Also, produce AD to H and join H and F



Here ABFH is a parallelogram
HF\parallel AB
And   \angle AFH = \angle FAB   …..(1)             {alternate interior angles}

AB = HF                                                 (opposite sides of a parallelogram)

AF = AF                                                 (Common Side)
\Rightarrow \triangle HAF\cong \triangle FAB  …..(2) {\ SAS Congruency}
Now, \angle HAF=\angle FAB …..(3) (EA is the bisector of \angle A)
\Rightarrow \angle AFH=\angle HAF        { using 1 and 3 }
So, HF = AH                   {Sides opposite to equal angles are equal}
 HF = AB = 10 cm
AH = HF = 10 cm
AD + DH = 10 cm
DH = 10 -AD \Rightarrow 10 -6
DH=4cm
Because FHDC is a parallelogram
  opposite sides are equal
DH = CF = 4 cm
Hence the answer is 4 cm

 

Posted by

infoexpert26

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